求满足2X^3-3X^2+(2-Y)X+Y-7=0的正整数X,Y的值

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求满足2X^3-3X^2+(2-Y)X+Y-7=0的正整数X,Y的值求满足2X^3-3X^2+(2-Y)X+Y-7=0的正整数X,Y的值求满足2X^3-3X^2+(2-Y)X+Y-7=0的正整数X,Y

求满足2X^3-3X^2+(2-Y)X+Y-7=0的正整数X,Y的值
求满足2X^3-3X^2+(2-Y)X+Y-7=0的正整数X,Y的值

求满足2X^3-3X^2+(2-Y)X+Y-7=0的正整数X,Y的值
若X=1,则原式不成立；
若X≠1,对原式化简,得Y=(2X^3-3X^2+2X-7)/(X-1)=
（2X^3-2X^2-X^2+X=X-1-6）/(X-1)=
2X^2-X+1-6/(X-1)
故X=2,Y=1
OR X=3,Y=13
OR X=4,Y=27
OR X=7,Y=91

由2X^3-3X^2+(2-Y)X+Y-7=0得：
(X-1)Y=2X^3-3X^2+2X-7
将X=1代入原等式，右边=-6
故X=1不符合题意
故由：(X-1)Y=2X^3-3X^2+2X-7
得：Y=(2X^2-X+1)+6/(X-1) 这里用了多项式的除法
若使Y为正整数，X只能取2,3,4,7
此时，Y值分别为7,16,29,92

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